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3 years ago

A 45-kg child sits on a 3.2-kg tire swing. What is the tension in the rope that hangs from a tree branch?

Physics

1 answer:

Aleksandr [31]3 years ago

7 0

Answer:

472.36 N

Explanation:

The total mass with the tire and child would be 45 kg + 3.2 kg = 48.2 kg

Ft = force tension

Fg = force of gravity

Fnet=0 (because child is motionless and not accelerating)

0 = Ft - Fg

Ft = Fg

Ft = m*g

Ft = 48.2 kg * 9.8 m/s^2

Ft = 472.36 N

Hope this helps!!

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When landing after a spectacular somersault, a 35.0 kg gymnast decelerates by pushing straight down on the mat. calculate the fo

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The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.

Now from Newton`s first law, the net force on gymnast,

$F_{net} =F-W=ma$

Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast ( $9\times g$ acceleration due to gravity)

Therefore,

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Substituting these values in above formula and calculate the force exerted by the gymnast,

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3 years ago

A ball rolls from rest at the top of a 15.5 m hill, downward to the right and back upward onto a 8.25 m hill. How fast is the ba

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3 years ago

Consider the following statement:

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The statement "The magnetic field of a magnet comes out of the north pole and goes into the south pole" is imprecise

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4 years ago

Find the electric field at a point midway between two charges of 30.0×10 power -9 and 60.0×10 power -9 separated by a distance o

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Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹ N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹ N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

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Eb = -(9.0 * 10⁹ N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb = -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

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3 years ago